# Tag Info

0

I guess that with projection you mean an orthogonal projection: $p^2=p=p^*$. For such a projection the image $im\, p$ and the kernel $ker \, p$ are closed subspaces of $H$. One has $H=im\, p\oplus ker\, p$, an othogonal sum, which means that each vector $x\in H$ can be presented uniquely as $x=y+z$, with $y\in im\, p$ and $z\in ker\, p$, i.e. $y\perp z$. ...

3

The assumption $p+q\le 1$ is equivalent to $$\forall x\in\mathcal{H},\qquad<(p+q)x,x>\le \Vert x\Vert^2$$ or equivalently $$\forall x\in\mathcal{H},\qquad \Vert p (x)\Vert^2+\Vert q(x)\Vert^2\le \Vert x\Vert^2$$ Applying this to $x=q(y)$ we get $$\forall y\in\mathcal{H},\qquad \Vert p (q(y))\Vert^2+\Vert q(y)\Vert^2\le \Vert q(y)\Vert^2$$ That is ...

0

The "strong operator topology on $H^*$" is just the same as the weak topology on $H$, so you're asking whether strong and weak convergence in $H$ is the same. No, it isn't. Standard example: an orthonormal sequence $u_n$ converges to $0$ weakly but does not converge strongly.

3

The first condition you will want to impose is that if $N$ is a null-set, then so is $g^{-1}(N)$. To see this, note that in general $\chi_{M} \circ g = \chi_{g^{-1}(M)}$, where $\chi_{M}$ is the characteristic function/indicator function of the set $M$. Hence, if there was a null-set $N \subset [a,b]$ such that $g^{-1}(N)$ is not a null-set, then you would ...

1

HINT: Use the change of variable formula to compare $$\int_I |f( s)|^2 d s= \int_I |f( \phi(t))|^2 | \phi'(t)|\, d t \ \text{and}\\ \int_I |f( \phi(t))|^2 d t \\$$

2

Two and a half hints: You can without loss of generality assume $z_0 = 0$ and $s = 1$, the general case differs only in book-keeping. Determine the inner product (and hence the norm) in terms of the coefficients of the Laurent series. You should see a connection with a familiar Hilbert space. Find what the absolute convergence of the Laurent series means ...

1

So does the author just mean that the basis is not a Hamel basis? Yes, precisely that; "a basis, in the sense of algebra" is a Hamel basis. I think that a total orthonormal sequence must be a Schauder basis Yes, that is correct, but a total orthonormal family need not be countable, in contrast to a sequence, and if you have an uncountable ...

4

In Hilbert space theory and Functional analysis the term 'basis' has a different definition than in linear algebra. For this reason one often encounters the term 'Hamel basis' for the latter and 'Hilbert space basis' in the first case. In the Hilbert space case the requirement is that each vector can be represented as in infinite sum, equality being ...

5

Take $\ell^2$ for example, i.e. the square-summable sequences of complex numbers with inner product $$\langle x,y\rangle = \sum_{n=1}^\infty x_n\overline{y_n}.$$ This has the countable orthonormal basis $$\{(1,0,0,\ldots), (0,1,0,\ldots), (0,0,1,0,\ldots),\ldots\}.$$ As $$\sum_{n=1}^\infty 2^{-n} = 1<\infty,$$ we have $$(2^{-1}, 2^{-2}, ... 1 Since the space is infinite dimensional so its basis cant be finite.Now every complete metric space is of 2nd category and by Baire Category theorem it cant be expressed as a countable union of Nowhere dense sets .can you take it from here? 0 The open balls of radius \frac{1}{2\sqrt2} around every point belonging the the orthonormal basis vectors are disjoint. A countable set can't intersect them all if there are uncountably many vector in you Hamel basis, so it is not dense, and then the space is not separable. So there is a problem with your set of hypothesis. What preceeds is conditional to ... 0 Here are no domain issues! (Caution that swapped version may give severe ones!) A formal calculation shows:$$P\psi\in\mathcal{D}(N):\quad\langle\varphi,NP\psi\rangle=\langle \varphi,NP\psi\rangle=\int\lambda\mathrm{d}\langle \varphi,E(\lambda)P\psi\rangle\quad(\varphi\in\mathcal{H})$$(Warning on algebraic properties of restricted spectral measures.) 0 Exploiting reducibility one gets:$$P\psi\in\mathcal{D}:\quad\int\lambda\mathrm{d}\langle\varphi,E(\lambda)P\psi\rangle=\langle\varphi,NP\psi\rangle=\langle P\varphi,NP\psi\rangle=\int\lambda\mathrm{d}\langle\varphi,PE(\lambda)P\psi\rangle\quad(\varphi\in\mathcal{H})$$Thus by uniqueness of spectral measures:$$PE(A)\equiv ...

0

There are no domain issues since: $$\|UE(A)U^*\psi\|=\|E(A)U^*\psi\|$$ But a formal calculation shows: $$U^*\psi\in\mathcal{D}(N):\quad\langle\varphi,UNU^*\psi\rangle=\int\lambda\mathrm{d}\langle U^*\varphi,E(\lambda)U^*\psi\rangle=\int\lambda\mathrm{d}\langle\varphi,UE(\lambda)U^*\psi\rangle\quad(\varphi\in\mathcal{H})$$ Concluding that the identity ...

0

Partial Isometry Before, it is a partial isometry: $$\|\Omega\varphi\|=\lim_{\sigma\to\infty}\|e^{i\sigma H}e^{-i\sigma H_0}P_\text{ac}(H_0)\varphi\|=\lim_{\sigma\to\infty}\|P_\text{ac}(H_0)\varphi\|=\|P_\text{ac}(H_0)\varphi\|$$ So, denote for shorthand: ...

1

First show that if $E$ is a spectral measure then so is $UE(\cdot)U^*$. Now note that for a spectral measure $UE(\cdot)U^*$ we have $$\left<f, \int \limits_{\sigma(N)}z \ UE(\mathrm{d}z)U^* g\right>= \int\limits_{\sigma(N)} z\ F_{f, g}(dz) \ \quad (f \in \mathcal{H}, g \in U \mathcal{D}),$$ where $F_{f, g}(\cdot):=\left<f, UE(\cdot)U^*g\right>$ ...

4

Fundamentally, "equipped with" just means "and". A Hilbert space is an object which consists of two components: a vector space, and an inner product on that vector space. This means that strictly speaking, a Hilbert space and a vector space are different objects: a Hilbert space is not a vector space, and a vector space is not a Hilbert space. But the ...

1

We can show that the sequence $Y_n=\sum_{i=1}^na_iX_i$ is a Cauchy sequence. Suppose that $n>m$. Then $$\operatorname E[Y_n-Y_m]^2=\operatorname E\bigg[\sum_{i=m+1}^na_iX_i\biggr]^2=\operatorname EX_1^2\sum_{i=m+1}^na_i^2$$ using the independence and identical distributions of $X_1,X_2,\ldots$. Since the series $\sum_{i=1}^\infty a_i^2$ converges, the ...

1

Let $m < n \in \mathbb{N}$. Then, by the independence of the random variables (Bienaymé formula), \begin{align*} \left\| \sum_{i=1}^n a_i X_i - \sum_{i=1}^m a_i X_i \right\|_{L^2}^2 &= \left\| \sum_{i=m+1}^n a_i X_i \right\|_{L^2}^2 \\ &= \sum_{i=m+1}^n a_i^2 \underbrace{\|X_i\|_{L^2}^2}_{\text{var}(X_i)=1} \\ &\stackrel{n,m \to ... 1 I alluded to a method in my comment, but I guess I'll leave it here as an answer. Let \delta=\inf_{c\in C}\|c-h\|. Suppose x,y\in C are such that \|x-h\|=\|y-h\|=\delta. Using the parallelogram identity, we have\|x-y\|^2=2\|x-h\|^2+2\|y-h\|^2-\|(x-h)+(y-h)\|^2=4\delta^2-4\|\frac{x+y}2-h\|^2.$$Using the fact that C is convex, \frac{x+y}2\in C so ... 0 For closed subspaces one has:$$Z=\overline{Z}\implies\mathcal{H}=Z\oplus_\perp Z^\perp$$For orthogonal decompositions one has:$$\mathcal{H}=Z\oplus_\perp Z'\implies Z'=Z^\perp$$Taking both together one gets:$$\overline{U}=\left(\overline{U}^\perp\right)^\perp=U^{\perp\perp}$$(This works even for the closure on the span of plain sets.) 2 Informally, let's consider a mathematical space, which we think of primarily as a set A of points (or primitive elements) on which are defined various operations (e.g. addition, scalar multiplication with a field, scalar product, etc.). We can say that A is "equipped" with these operations. That is, whenever we mention A, we have in mind these ... -1 Given the Fock space \mathcal{F}(\mathcal{h}). Consider the second quantization:$$\Gamma(e^{ith})=e^{it\mathrm{d}\Gamma(h)}$$It acts continuous on the CAR-algebra:$$\quad\tau^t[a(\eta)]=a(e^{ith}\eta)\stackrel{t\to0}{\to}a(\eta)$$(Remember that this won't work for the Weyl algebra.) 0 Each Hilbert space (X, <\cdot,\cdot>) is also a normed space (X, ||\cdot||) with norm ||x|| := \sqrt {<x,x>}, x\in X. Concerning the opposite direction: If (X, ||\cdot||) is a normed vector space, then its norm must satisfy the parallelogram law ||x+y||^2 + ||x-y||^2=2||x||^2 + 2||y||^2 in order to derive from an inner product. Not ... 11 I think the best way of explaining what it means to be equipped with something is by showing how the term is used. Mathematics has a very precise formalism, but in its pure form it is mostly not appropriate for daily use because then even the simplest problems would become very hard. Equipping a mathematical object with additional structures is one of the ... 1 The underlying idea is to have a space of functions instead of a space of points. What do you need to make this work? It turns out that the most important properties (ie topological properties) all follows from the notion of distance. It is intuitive what is the distance between two points. But what is the distance between two functions? If you define ... 4 f_n \in L^2: No, in general this is not true. Consider for example \Omega = (0,1) endowed with the Lebesgue measure and$$f_n(x) := \frac{1}{n} x^{-1/4}.$$Then f_n \to f:=0 in L^2 and therefore f \cdot g=0 \in L^2 for any g \in L^2. On the other hand, if we choose g(x) := x^{-1/4} \in L^2, then f_n \cdot g \notin L^2 for all n \in ... 0 Hint: If x\not\in \overline{U}, then by Hilbert projection theorem we have x=y+z with y\in \overline{U} and 0\ne z\perp \overline{U}. Since z\ne 0 one has$$ \langle x,z\rangle=\langle y+z,z\rangle=\| z\|^2\ne 0. $$Hence z\in U^\perp does not annihilate x and therefore x\not\in U^{\perp\perp}. 0 Evan's hint: For all v\in H$$ \int_0^T (v,u_k)\leq C||v|| T. $$Since L^2(0,T;H) is Hilbert, the assumption u_k \rightarrow u weakly in L^2(0,T;H) reads $$\int_0^T (v,u(t)) dt=\lim_{k\to\infty}\int_0^T(v,u_k(t)), \,\,\forall v\in L^2(0,T;H)$$ Consider the particular case, v\in L^2(0,T;H) as v= w with w\in H ... 0 It seems the following. Suppose that there exist points x and y in a normed space X such that \|x\|\ge 1, y\in C and$$\|x-y\|<\|x-P_C(x)\|.$$Then Triangle inequality implies$$\|x\|\le \|x-y\|+\|y\|<\|x-P_C(x)\|+1=\|x-P_C(x)\|+\|P_C(x)\| =\|x\|,$$a contradiction. 0 If u^* is a partial isometry, then your claim is correct, but in general Im u^*u is not closed. To prove your claim in general we should show \overline {Im u^*} = \overline {Im (u^*u)} . Reduced it to show \ker u = \ker u^*u. Clearly \ker u\subset \ker u^*u. Conversely we can decompose H' = \overline{Im(u)} + \ker u^*. Suppose \xi \in \ker ... 46 The word "equipped" keeps notational pandemonium from breaking loose. For instance, if you were to be a bit more formal, you'd say A Hilbert space is a pair (V, \left<\cdot,\cdot \right>), where V is a vector space and \left<\cdot,\cdot \right>\colon V\times V \to \mathbb{C} is an inner product. Additionally, all Cauchy sequences in ... 19 Generally, when we use the word "equipped", we mean not that it is possible to do an operation, but we have one in mind. That is on a vector space like \mathbb R^2, we could think of various inner products that would suffice. For instance the following two operations are both inner products:$$(x_1,x_2)\cdot(y_1,y_2)=x_1y_1+x_2y_2$$... 9 Well, sometimes we actually want to work with spaces were as few operations as possible are allowed. Often in math, we want to find the most general environment where a property holds, so we want to work in spaces with as few properties as possible. It's interesting to see just what we can do without needing to use an inner product, or a metric, or the lack ... 0 First note that the inverse has full domain:$$\int_\mathbb{R}\left|\frac{1}{1+e^{-\beta\lambda}}\right|^2\mathrm{d}\nu_\zeta(\lambda)\leq\int_\mathbb{R}1\mathrm{d}\nu_\zeta(\lambda)=\|\zeta\|^2<\infty$$So one has a common domain:$$\mathcal{D}\left(H^k\circ\frac{1}{1+e^{-\beta H}}\right)=\mathcal{D}\left(\frac{H^k}{1+e^{-\beta ...

-2

It is true. It follows the condition of optimality for optimization problem. Proof: Consider $F(y) = \left<x-y,x-y\right>$. The problem $$F(y)\to\min,\ y\in M$$ has a unique solution, $P(x)$. Since $F$ is a convex function and $M$ is a convex set, then the following holds: $$\left<\nabla F(P(x)), v - P(x)\right> \geqslant 0\text{ for all ... 2 If you have an integral transform Uf = \int_{-\infty}^{\infty} P(s,x)f(x)dx with an inverse integral transform Vg = \int_{-\infty}^{\infty} Q(x,s)g(s)ds, then you'll have something similar to the Fourier transform \delta-function behavior because$$ \begin{align} f(y) & =(VUf)(y) \\ & ...

1

The best way to see what can go wrong is to find a counterexample. The easiest counterexample I can think of here is $\ell^2$. Let $c_0 \subset \ell^2$ denote the set of sequences $(x_j)$ such that there exists an $n \in N$ such that $x_j = 0$ whenever $j \geq N$. That is, let $c_0$ denote the set of all "terminating" sequences. What is $(c_0)^\perp$? Is ...

0

$$(\sqrt{\lambda}||x||-\frac{1}{2\sqrt{\lambda}}||y||)^2 \ge 0 \iff \lambda||x||^2+\frac{1}{4\lambda} ||y||^2 \ge ||x|| \cdot ||y||$$ but $$||x|| \cdot ||y|| \ge |\langle x, y \rangle|$$

0

Note that by AM-GM inequality we have $$\lambda \|x\|^2+(1/4\lambda)\|y\|^2\geq \|x\|\|y\|$$ The desired result now follows from the Cauchy-Schwarz inequality.

1

1) $u(x)\equiv0$ is in $K$. 2) If $f_n\to f$ in $L^2$, there is a subsequence $f_{n_k}$ converging to $f$ almost everywhere. Since $|f_n(x)|\le h(x)$ a.e, it follows that $|f(x)|\le h(x)$ a.e. and $f\in K$. 3) The condition you state for $P_Kf$ is incorrect (it would be OK if $K$ where a subspace.) It should be $$\langle ... 1 Let me also give a hint on how to start:$$\|v -P_K f \|^2 = \langle v -f +f -P_K f, v- P_K f\rangle = \langle v -f, v - P_K f\rangle + \langle f -P_K f, v - P_K f\rangle$$With my hint, i.e. \langle f -P_K f, v -P_K f\rangle\leq 0 (at least in a real Hilbert-space, otherwise this only holds for the real part of the expression) applied to the second term ... 1 \forall v,u\in H it holds: ||v+u||^2=||u||^2+||v||^2+2<u,v> (*), then: ||v-u||^2=||v-f+f-u||^2= =||(v-f)+(f-u)||^2=^{(*)} =||u-f||^2+||v-f||^2+2<v-f,f-u>=(1) Now we have: <v-f,f-u>=<v-u+u-f,f-u>=<v-u,f-u>-||u-f||^2, therefore: (1)=||u-f||^2+||v-f||^2+2(<v-u,f-u>-||u-f||^2)= ... 1 You're absolutely right. You can just as easily define S^\bot for any subset S of your Hilbert space V.$$S^\bot = \{ v \in V : \langle v \vert w \rangle = 0 \text{ for all } w \in S \}$$You can also check that, no matter the structure of S, S^\bot is a closed subspace, and (S^\bot)^\bot is the smallest closed subspace containing S. 1 Suppose A is selfadjoint with a complete orthonormal basis of eigenvectors \{ e_{n}\}_{n=1}^{\infty} and corresponding eigenvalues \{ \lambda_{n}\}_{n=1}^{\infty} that tend to \infty in absolute value. Then$$ f \in \mathcal{D}(A^{N}) \iff \sum _{n=1}^{\infty}\lambda_{n}^{2N}|(f,e_{n})|^{2} < \infty.  The behavior of the ordinary Fourier ...

1

Let $T_n(x) = \sum_{k=1}^n {1 \over k} x_k e_{k+1}$. $T_n$ has finite rank, so it is compact. We have $\|T-T_n\| \le {1 \over n+1}$, hence $T_n \to T$ and so $T$ is compact.

1

Hint: You might show the image of the unit ball is totally bounded. Given $\epsilon > 0$, take $N$ large enough, and cover the projection of the image of the unit ball onto the first $N$ coordinates by small balls ...

0

If $e_j$ is an orthonormal basis, then you will always have equality, by definition. The only way to achieve inequality is if $e_j$ is orthonormal but not a basis, for example as you did by just taking even terms.

0

I tend to disagree with the arguments given by Stefan Rauh. First of all $l^\infty$ is not a Hilbert Space. However every discrete set (e.g. a countable set of numbers or a set of files or a set of Git repositories) can always be interpreted as a 0-manifold. If these manifoldes are to be interpreted as a submanifold of a Hilbert space there must be some ...

2

Take a basis $\{v_i\}$ of $N$. In particular $v_i=Qv_i=UU^* v_i$. We can show that $U^*v_i$ is a basis of $M$. In fact, $U^*v_i=U^*Qv_i=U^*UU^*v_i=PU^*v_i$, so $U^*v_i\in M$. If $0=\sum a_iU^*v_i$. Then $0=\sum a_iUU^*v_i=\sum a_iv_i$. Therefore $a_i=0$, i.e. $U^*v_i$ are linearly independent. If $x\in M$, then $x=Px=U^*Ux$, then $Ux=UU^*Ux=QUx$. This ...

Top 50 recent answers are included